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For a visual reference of what I mean by semi-cylindrical plates check the diagram below.

semi-cylindrical capacitor

EDIT: The above diagram is not what I intend to implement. It is just to show anyone trying to answer the problem what I mean by semi-cylindrical plates.

How would one go about deriving the formula for such a capacitor? I could start with something but don't know how to proceed further. Here is what I could come up with.

top view semi cylindrical capacitor plates

If we start with the equation of parallel plate capacitor $$ C = \dfrac{\epsilon A}{d} $$

and replace d = f(x), x being the angle shown in the diagram. Then by the formula of chord length $$ f(x) = 2rsin(x) $$

this makes the original equation $$ C = \dfrac{\epsilon A}{2rsin(x)} $$

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    \$\begingroup\$ From Wikipedia "... two thin parallel conductive plates each with an area of A separated by a uniform gap of thickness d. It is assumed the gap d is much smaller than the dimensions of the plates". This is not applicable in the above case. The expression may have to be derived from first principles. \$\endgroup\$
    – AJN
    Jul 15, 2020 at 15:06
  • \$\begingroup\$ You will want to increase X alot, so that it measures in the middle more than the edge. So just measure it and calibrate it. Since water C, Dk= 80, moisture content and fertilizer (R=?) it is better to use a current source signal then measure voltage as impedance. \$\endgroup\$ Jul 15, 2020 at 15:09
  • \$\begingroup\$ If x is increased, the capacitance is expected to decrease. Will it be large enough to measure ? OP might as well do away with the cylindrical shape and stick with two parallel plates ? \$\endgroup\$
    – AJN
    Jul 15, 2020 at 15:13
  • \$\begingroup\$ @AJN regarding your first comment, this presents a further step I might try. I know that the assumed gap d comes from equation dV = E \dot dS which when integrated over the limits of the gap would evaluate to d. So if I were to re-write the chord length formula as a derivative of f(x) w.r.t to x. Then I could substitute it in the voltage equation and derive the capacitance formula. Would this approach be right? I am wondering what the limits of the integration would be in such a case. \$\endgroup\$ Jul 15, 2020 at 16:08
  • \$\begingroup\$ if I were to re-write the chord length formula as a derivative of f(x). I am not sure if it would be the way to go. \$\endgroup\$
    – AJN
    Jul 15, 2020 at 16:17

2 Answers 2

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Have you considered using a conformal mapping technique? This hopefully will give you a closed form expression for the capacitance per unit of height of the cylinders neglecting the fringe fields at the end of the cylinders.

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  • \$\begingroup\$ I do not know about conformal mapping technique? Could you please elaborate? \$\endgroup\$ Jul 15, 2020 at 16:01
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    \$\begingroup\$ it is a method used for finding the dc Electrical characteristics of 2-dimensional structures or equivalently, structures that are infinite in length and uniform across their infinite extent. It uses a map (function in the complex plane) to transform a geometry in a simpler one where one can easily derive an analytical expression. For example, in your case, perhaps you could find a map to transform the cylinders into parallel plates, in which case the derivation is trivial. The method has been used to find the quasi-TEM characteristics of Coplanar waveguide transmission lines etc \$\endgroup\$ Jul 15, 2020 at 16:20
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You could divide the circular plates into small segments, each of which approximates a parallel plate capacitor, for which you can easily find the capacitance. The first segment would be the two arcs closest to the center. Then rotate the geometry a small amount clockwise, and you still have two tiny parallel plates. Continue around in both directions and sum the C values. Of course this is an elementary calculus exercise.

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  • \$\begingroup\$ I don't think this will work. The distance between the small segments will be much larger than their width, so the parallel plate approximation will be grossly incorrect. \$\endgroup\$ Jul 15, 2020 at 20:48
  • \$\begingroup\$ @ElliotAlderson Each small segment has a corresponding small segment that is diametrically opposite, so always the same distance apart. So it works fine. If you carry out the calculation you will find that the capacitance is given by (epsilon) (theta)(depth), where theta is the angle subtended by the semicircular plates. Interestingly, this result is independent of the diameter. \$\endgroup\$ Jul 16, 2020 at 17:41
  • \$\begingroup\$ But the width of the segments is much smaller than the distance between them so you can no longer use parallel plate model...you have to account for fringing at the edges. \$\endgroup\$ Jul 16, 2020 at 18:03
  • \$\begingroup\$ @ElliotAlderson You could decompose a parallel plate capacitor whose width is much larger than the distance between the plates into exactly the same infinitesimal segments and derive the usual result. So it seems to me that the fringing effect is not important until you reach the edges of the plates. \$\endgroup\$ Jul 16, 2020 at 18:31

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